Prove that $h(y)=\int_{y}^{y+1}g(x)dx$ is integrable and $\int_{\mathbb{R}}h\lambda_1=\int_{\mathbb{R}}g\lambda_1$, given that $g$ is integrable.

Question

We have $g:\mathbb{R}\rightarrow\mathbb{R}$, integrable on $\mathbb{R}$

Let's define: $h(y)=\int_{y}^{y+1}g(x)dx$

Prove that $h(y)$ is integrable and $\int_{\mathbb{R}}h\lambda_1=\int_{\mathbb{R}}g\lambda_1$

I have no idea how to do it.

Answer 1

$h$ is integrable if $\int_{\Bbb R} |h(y)| dy < \infty$ so let's have a look und use Fubini to get:

$$ \begin{align*} \int_{\Bbb R} |h(y)| dy &= \int_{\Bbb R} \left|\int_{y}^{y+1} g(x) dx\right| dy\\ &\le \int_{\Bbb R} \int_{y}^{y+1} |g(x)|\; dx\; dy \\ &= \int_{\Bbb R} |g(x)| \int_{x-1}^{x}1\; dy\; dx \\ &= \int_{\Bbb R} |g(x)| dx < \infty \end{align*}$$

So h is integrable and we can recap the same steps without the absolute value and the estimation to get $$\int_{\Bbb R} h(y) dy = \int_{\Bbb R} g(x) dx$$